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lutik1710 [3]
3 years ago
6

An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductan

ce of L = 0.2 henrys, a resistor of R = 5 ohms and a capacitor of capcitance C = 0.043 farads. What is the amplitude of the current I?
Engineering
1 answer:
german3 years ago
4 0

Answer:

Explanation:

we have given E(t)=120 sin(12t)

R=5 ohm

L=0.2 H

ω=12 ( from expression of E)

X_L=0.2\times 12=2.4 ohm

X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm

Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}

Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}

=5.021 ohm

so amplitude of current =  \frac{v}{z}=\frac{120}{5.021}=23.89

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A steam pipe passes through a chemical plant, where wind passes in cross-flow over the outside of the pipe. The steam is saturat
valina [46]

Answer:

a) the rate of heat transfer from the pipe to the air is 23.866 watts

b) YES, the rate of heat transfer changes to 3518.61 watt

Explanation:

Given that:

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Bulk fluid temperature of 27°C

we know that

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hD/k = 0.028 (Dvp / u)^0.8 (Cpu / k)^0.33

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([0.0675 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

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a)

Now to find the rate of heat transfer Q

Q = hAΔT

Q = 0.0414 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 23.866 watts

therefore the rate of heat transfer from the pipe to the air is 23.866 watts

b)

Now the flow direction changes to parallel flow, then

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([34.7 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 6.1036 w/m².k

so from the steam table, saturated steam at 17.70 bar, temperature of steam will be 105.383°C

so to find the rate of heat transfer Q

Q = hAΔT

Q = 6.1036 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 3518.61 watt

Therefore the rate of heat transfer changes to 3518.61 watt

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